|
|
|
|
|
|
|
|
|
|
|
|
$$$9 t^{2} + 4$$$的导数
您的输入
求$$$\frac{d}{dt} \left(9 t^{2} + 4\right)$$$。
解答
和/差的导数等于导数的和/差：
$${\color{red}\left(\frac{d}{dt} \left(9 t^{2} + 4\right)\right)} = {\color{red}\left(\frac{d}{dt} \left(9 t^{2}\right) + \frac{d}{dt} \left(4\right)\right)}$$常数的导数是$$$0$$$:
$${\color{red}\left(\frac{d}{dt} \left(4\right)\right)} + \frac{d}{dt} \left(9 t^{2}\right) = {\color{red}\left(0\right)} + \frac{d}{dt} \left(9 t^{2}\right)$$对 $$$c = 9$$$ 和 $$$f{\left(t \right)} = t^{2}$$$ 应用常数倍法则 $$$\frac{d}{dt} \left(c f{\left(t \right)}\right) = c \frac{d}{dt} \left(f{\left(t \right)}\right)$$$：
$${\color{red}\left(\frac{d}{dt} \left(9 t^{2}\right)\right)} = {\color{red}\left(9 \frac{d}{dt} \left(t^{2}\right)\right)}$$应用幂次法则 $$$\frac{d}{dt} \left(t^{n}\right) = n t^{n - 1}$$$，其中 $$$n = 2$$$:
$$9 {\color{red}\left(\frac{d}{dt} \left(t^{2}\right)\right)} = 9 {\color{red}\left(2 t\right)}$$因此，$$$\frac{d}{dt} \left(9 t^{2} + 4\right) = 18 t$$$。
答案
$$$\frac{d}{dt} \left(9 t^{2} + 4\right) = 18 t$$$A
Please try a new game Rotatly